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Consider the following equation for $t > 1$:

$$(t + \sqrt{t^2 - 1})^{x^2 - 2x} + (t - \sqrt{t^2 - 1})^{x^2 - 2x} = 2t$$

If we let $u = t + \sqrt{t^2 - 1}$ then $\frac{1}{u} = t - \sqrt{t^2 - 1}$ and the equation reduces to $$u^{x^2 - 2x} + \frac{1}{u^{x^2 - 2x}} = u + \frac{1}{u}$$

Equating $x^2 - 2x$ to $1$ and $-1$ gives us three roots.

How can we prove that there are no more roots or find those if there are?

Afntu
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Shiv
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1 Answers1

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Denote $a = u^{x^2-2x}$, then $$\begin{align} &\iff a-u+\frac{1}{a} - \frac{1}{u} =0 \\ &\iff (a-u)\left(1-\frac{1}{au}\right)=0 \\ &\iff \cases{a =u\\ a = \frac{1}{u}} \\ &\iff \cases{x^2 -2x =1\\ x^2 - 2x = -1} \\ &\iff \color{red}{x \in \left\{1, 1\pm \sqrt{2} \right\}} \end{align}$$

NN2
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